On the basis of the following thermochemical data (Delta_(f)G^(@)H^(+)(aq) = 0) H_(2)O(l) to H^(+)(aq) + OH^(-)(aq), DeltaH = 57.32 kJ (ii) H_(2)(g) + 1/2 to H_(@)O(l), DeltaH = -286.2 kJ The value of enthalpy of formation of OH^(-) at 25^(@)C is

On the basis of the following thermochemical data (Delta_(f)G^(@)H^(+)

1.	On the basis of the following thermochemical data (Delta_(f)G^(@)H^(+)(aq) = 0) H_(2)O(l) to H^(+)(aq) + OH^(-)(aq), DeltaH = 57.32 kJ (ii) H_(2)(g) + 1/2 to H_(@)O(l), DeltaH = -286.2 kJ The value of enthalpy of formation of OH^(-) at 25^(@)C is
Answer» `-22.88` kJ `-228.88` kJ `+228.88` kJ `-343.52` Kj Solution :`H_(2)(g) + 1/2O_(2)(g) to H^(+)(AQ) + OH^(-)(aq)DeltaH_(1) = ?` `H_(2)O(l) to H^(+)(aq) + OH^(-)(aq)DeltaH_(2) = 57.32 kJ` ..(i) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l)DeltaH_(3) = -286.2` kJ ...(II) `DeltaH_(1)` will be OBTAINED by ADDING (i) and (ii) ` = 57.32 + (-286.2) = -228.88 kJ`